In the figure,the magnetic needle has a magne | vedclass

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(A) The time period of oscillation $T$ is given by the total time divided by the number of oscillations: $T = \frac{6.70 \; s}{10} = 0.67 \; s$.The fo

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$0.01$

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(A) The time period of oscillation $T$ is given by the total time divided by the number of oscillations: $T = \frac{6.70 \; s}{10} = 0.67 \; s$.The formula for the time period of a magnetic needle oscillating in a magnetic field is $T = 2\pi \sqrt{\frac{I}{mB}}$.Rearranging this formula to solve for the magnetic field $B$,we get: $B = \frac{4\pi^2 I}{m T^2}$.Substituting the given values: $B = \frac{4 	imes (3.14)^2 	imes 7.5 	imes 10^{-6}}{(6.7 	imes 10^{-2}) 	imes (0.67)^2}$.Calculating the denominator: $(6.7 	imes 10^{-2}) 	imes 0.4489 \approx 0.0300763$.Calculating the numerator: $4 	imes 9.8596 	imes 7.5 	imes 10^{-6} \approx 0.000295788$.Thus,$B = \frac{0.000295788}{0.0300763} \approx 0.00983 \; T$,which is approximately $0.01 \; T$.

In the figure,the magnetic needle has a magnetic moment $m = 6.7 \times 10^{-2} \; A \cdot m^2$ and a moment of inertia $I = 7.5 \times 10^{-6} \; kg \cdot m^2$. It performs $10$ complete oscillations in $6.70 \; s$. What is the magnitude of the magnetic field $B$ (in $; T$)?

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